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Question Number 143287 by enter last updated on 12/Jun/21

Answered by Huy last updated on 12/Jun/21

$$\mathrm{Dieu}\mathrm{kien}:\mathrm{x}⩾-1$$$$\mathrm{Phuong}\mathrm{trinh}\mathrm{tuong}\mathrm{duong}:$$$$2({\mathrm{x}}^2+2)=5\sqrt{(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)}$$$$<=>2({\mathrm{x}}^2+2)=5\sqrt{(\mathrm{x}+1)[({\mathrm{x}}^2+2)-(\mathrm{x}+1)]}$$$$\mathrm{Dat}\mathrm{a}={\mathrm{x}}^2+2\mathrm{va}\mathrm{b}=\mathrm{x}+1(\mathrm{a}>\mathrm{b}⩾0)$$$$=>2\mathrm{a}=5\sqrt{\mathrm{b}(\mathrm{a}-\mathrm{b})}$$$$<=>{4\mathrm{a}}^2=25\mathrm{b}(\mathrm{a}-\mathrm{b})$$$$<=>{4\mathrm{a}}^2-25\mathrm{ab}+{25\mathrm{b}}^2=0$$$$\bullet \mathrm{Voi}\mathrm{b}=0=>\mathrm{a}=0\left(\mathrm{loai}\right)$$$$\bullet \mathrm{Voi}\mathrm{b}\ne 0,\mathrm{chia}\mathrm{hai}\mathrm{ve}\mathrm{cho}{\mathrm{b}}^2\mathrm{ta}\mathrm{duoc}$$$$4{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^2-25\frac{\mathrm{a}}{\mathrm{b}}+25=0$$$$<=>\frac{\mathrm{a}}{\mathrm{b}}=5\mathrm{hoac}\frac{\mathrm{a}}{\mathrm{b}}=\frac{5}{4}$$$$\bullet \mathrm{Truong}\mathrm{hop}\frac{\mathrm{a}}{\mathrm{b}}=5:$$$$\mathrm{a}=5\mathrm{b}<=>{\mathrm{x}}^2+2=5(\mathrm{x}+1)<=>\mathrm{x}=\frac{5+\sqrt{37}}{2}$$$$\bullet \mathrm{Truong}\mathrm{hop}\frac{\mathrm{a}}{\mathrm{b}}=\frac{5}{4}:$$$$4\mathrm{a}=5\mathrm{b}<=>4({\mathrm{x}}^2+2)=5(\mathrm{x}+1)\left(\mathrm{vo}\mathrm{nghiem}\right)$$$$\mathrm{Vay}\mathrm{x}=\frac{5+\sqrt{37}}{2}$$

Answered by MJS_new last updated on 12/Jun/21

$$2(x^2+2)=5\sqrt{x^3+1}$$$$4{(x^2+2)}^2=25(x^3+1)$$$$x^4-\frac{25}{4}{x}^3+4x^2-\frac{9}{4}=0$$$$(x^2-5x-3)(x^2-\frac{5}{4}x+\frac{3}{4})=0$$$$\Rightarrow$$$$x_{1,2}=\frac{5}{2}\pm \frac{\sqrt{37}}{2}$$$$x_{3,4}=\frac{5}{8}\pm \frac{\sqrt{23}}{8}\mathrm{i}$$$$\mathrm{testing}\mathrm{all}\mathrm{solutions}\mathrm{in}\mathrm{given}\mathrm{equation}\Rightarrow$$$$\mathrm{all}\mathrm{solutions}\mathrm{are}\mathrm{valid}$$

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